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You need to get it in the form (x-h)^2 + (y-k)^2 = r^2 to show that it is a circle... otherwise it could be an ellipse or another conic section.

The oblique asymptote is y=3x. Basically when you divide it out using long division the remainder part approaches zero as x approaches infinity. Thus it will approach the other part. The question is badly worded I think though...

Possible rational roots are p/q where p|4 and q|1...

To me there's no difference between the supposed 'strong' and 'weak' variations of induction...

I found Harder 3U the hardest and most annoying. My favourite was Mechanics. It was also the easiest for me (especially Circular Motion). Then the others were no different to me.

:rolleyes: Make up your mind Anna.

Lmao, "Phil for humanity"... Yes. But according to our friend Anna's 'theory', it would've been 1 :rolleyes:

Poor students...:uhoh:

Yeah PM me. I'll send you mine.

Open your eyes:

undefined is like 2/0 (which approaches infinity). Mathematically indeterminable is described by the 4 cases above.

Perhaps. But it's be more specific if you called the four cases above as being "mathematically indeterminable".

Ohhh... so BASIC questions huh... Then find: lim_{x \to \infty}\frac{x^2+x+1}{5x^2+10x+90} Look in the Coroneos 2U book. There's a good explanation there in the limits chapter.

Hmm yes. Then what value does lnx/x approach as x-->infinity? According to you it approaches 1, when it infact doesn't. Furthermore, you could get a question asking you to graph this. You would seriously mislead students in these types of questions if you told them that infinity/infinity=1 to...

I dunno if you'd call 0/0 undefined. It's more mathematically indeterminable. I think... Undefined would be something like 2/0

Nope. They're both mathematically indeterminable (or meaningless). See above for the four cases of an expression being mathematically indeterminable. The sad thing for our friend Anna is that she's already done 4U maths, doing uni maths and is tutoring students. Once I explained to...

You let it equal 1/2 for the required expression.

A joke? :S No you were dead serious. Oh, and neg rep to me is like infinity/infinity , 0/0, infinity - infinity or 0*infinity. ie. Mathematically indeterminable, or more briefly MEANINGLESS ;) :D Dw Anna I know who you are, quite literally xD

cos3\theta = 4cos^3\theta-3cos\theta \\ Let \ cos\3\theta= 1/2 \ \3\theta = \frac{\pi}3 , \ \frac{5\pi}{3} \ or \ \frac{7\pi}{3} \\ \theta = \frac{\pi}9 , \ \frac{5\pi}{9} \ or \ \frac{7\pi}{9} \\ Thus \ 4cos^3\theta - 3cos\theta = \frac{1}{2} \text{ will have the same roots} \\ Let \...

Yes, does infinity/infinity --> 1 now does it? o.O